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Chapter 4: Problem 21

Headway on four-lane highways: When traffic is flowing on a highway, the headway is the average time between vehicles. On four-lane highways, the probability \(P\) that the headway is at least \(t\) seconds is given to a good degree of accuracy \({ }^{6}\) by $$ P=e^{-q t}, $$ a. On a four-lane highway carrying an average of 500 vehicles per hour in one direction, what is the probability that the headway is at least 15 seconds? (Note: 500 vehicles per hour is \(\frac{500}{3600}=0.14\) vehicle per second.) b. On a four-lane highway carrying an average of 500 vehicles per hour, what is the decay factor for the probability that headways are at least \(t\) seconds? Reminder: An important law of exponents tells us that \(a^{b c}=\left(a^{b}\right)^{c}\). where \(q\) is the average number of vehicles per second traveling one way on the highway.

Short Answer

Expert verified
a. Probability is approximately 0.1225. b. The decay factor is 0.14.

Step by step solution

01

Understanding the Problem

The headway on a highway is the average time between vehicles. We need to calculate two things using the headway probability function given by \( P = e^{-qt} \), where \( q \) is the average number of vehicles per second, and \( t \) is the time in seconds.
02

Convert Average Vehicles Per Hour to Per Second

Given that the highway has an average of 500 vehicles per hour traveling in one direction, convert this to vehicles per second by dividing by the number of seconds in an hour: \( q = \frac{500}{3600} \approx 0.14 \) vehicles per second.
03

Calculate Probability for Part a

To find the probability that the headway \( t \) is at least 15 seconds, plug \( q = 0.14 \) and \( t = 15 \) into the formula: \( P = e^{-0.14 \times 15} \). Simplify the exponent: \( -0.14 \times 15 = -2.1 \). Therefore, \( P = e^{-2.1} \). Using a calculator, \( e^{-2.1} \approx 0.1225 \).
04

Determine Decay Factor for Part b

The decay factor in the formula \( P = e^{-qt} \) is \( q \), which is the rate at which vehicles pass a point per second. Since we already calculated \( q = 0.14 \) vehicles per second, this is the decay factor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Decay
The concept of exponential decay is crucial in various fields, including traffic flow analysis. It's a mathematical function that describes how a quantity decreases rapidly at first and then more slowly over time. In the context of traffic on a four-lane highway, this is described by the formula \( P = e^{-qt} \). Here, \( P \) represents the probability that the headway, or the interval between vehicles, meets or exceeds a certain time \( t \).
The term \( e^{-qt} \) involves the natural base \( e \), which is approximately 2.71828. When multiplied by a negative exponent, it reflects a form of decay 鈥� specifically, the chances decline as the headway time increases.
  • \( P \) is the probability that cars are spaced at least \( t \) seconds apart.
  • \( q \) is the decay constant, representing the rate of vehicle passage per second.
  • \( t \) is the headway time in seconds.
This exponential decay applies not only to headway but to any process where something decreases quickly then steadies. Understanding this pattern is key in analyzing how likely large headways are on busy highways.
Traffic Flow Analysis
Traffic flow analysis is a method used to understand how vehicles move along roadways. It involves the study of traffic patterns, speeds, and densities to optimize the movement of vehicles and reduce congestion. In our problem, traffic flow is assessed on a four-lane highway where headway probability helps to gauge traffic conditions.
A key aspect of analyzing traffic flow is understanding headway, which is the average time interval between vehicles. By calculating the headway, transportation planners can estimate the level of service on the highway. High headway values suggest lighter traffic, while lower values signify more congestion.
  • Analysis helps in planning road expansions or adjustments.
  • It's also essential for setting traffic lights and managing peak-hour traffic.
  • Applications include designing routes and predicting future traffic patterns.
By analyzing these aspects, planners aim to improve the efficiency of the road network, minimize travel times, and enhance overall road safety.
Rate of Vehicles
The rate of vehicles refers to the average number of vehicles passing a given point per unit of time, often expressed in vehicles per second or per hour. In this exercise, 500 vehicles per hour equates to a vehicle passage rate of approximately 0.14 vehicles per second (\[ q = \frac{500}{3600} \approx 0.14 \]). This number is pivotal in calculating the headway probability and understanding the decay factor.
The vehicle rate directly influences the headway since a higher rate suggests more frequent vehicle passage, reducing average headway. On highways, this rate helps determine road capacity and inform decisions regarding traffic management.
  • Higher rates (large \( q \)) result in shorter time intervals between cars.
  • Lower rates suggest less congestion and longer headways.
  • This rate provides insights crucial for urban transportation planning.
Overall, understanding the rate of vehicles is integral in adjusting infrastructure, optimizing travel times, and predicting traffic conditions effectively. This is central to both micro-level traffic studies and broader transportation policy developments.

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Most popular questions from this chapter

Stochastic population growth: Many populations are appropriately modeled by an exponential function, at least for a limited period of time. But there are many factors contributing to the growth of any population, and many of them depend on chance. There are a number of ways to produce stochastic models. We consider one that is an illustration of a simple Monte Carlo method. We want to model a population that is initially 500 and grows at an average rate of 2% per year. To do this we make a table of values for population according to the following procedure. The first entry in the table is for time t = 0, and it records the initial value 500. To get the entry corresponding to t = 1, we roll a die and change the population according to the face that appears, using the following rule. $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Face } \\ \text { appearing } \end{array} & \begin{array}{c} \text { Population } \\ \text { change } \end{array} \\ \hline 1 & \text { Down 2\% } \\ \hline 2 & \text { Down } 1 \% \\ \hline 3 & \text { No change } \\ \hline 4 & \text { Up 2\% } \\ \hline 5 & \text { Up 4\% } \\ \hline 6 & \text { Up 9\% } \\ \hline \end{array} $$ To get the entry corresponding to \(t=2\), we roll a die and change the population from \(t=1\), again using the above rule. This procedure is then followed for \(t=3\), and so on. a. Using this procedure, make a table recording the population values for years 0 through 10 . b. Plot the data points from your table and the exponential model on the same screen. Comment on the level of agreement.

Growth in length of haddock: A study by Riatt showed that the maximum length a haddock could be expected to grow is about 53 centimeters. Let D = D(t) denote the difference between 53 centimeters and the length at age t years. The table below gives experimentally collected values for D. $$ \begin{array}{|c|c|} \hline \text { Age } t & \text { Difference } D \\ \hline 2 & 28.2 \\ \hline 5 & 16.1 \\ \hline 7 & 9.5 \\ \hline 13 & 3.3 \\ \hline 19 & 1.0 \\ \hline \end{array} $$ a. Find an exponential model of \(D\) as a function of \(t\). b. Let \(L=L(t)\) denote the length in centimeters of a haddock at age \(t\) years. Find a model for \(L\) as a function of \(t\). c. Plot the graph of the experimentally gathered data for the length \(L\) at ages \(2,5,7,13\), and 19 years along with the graph of the model you made for \(L\). Does this graph show that the 5 year-old haddock is a bit shorter or a bit longer than would be expected? d. A fisherman has caught a haddock that measures 41 centimeters. What is the approximate age of the haddock?

Long-term population growth: Although exponential growth can often be used to model population growth accurately for some periods of time, there are inevitably, in the long term, limiting factors that make purely exponential models inaccurate. If the U.S. population had continued to grow by \(3 \%\) each year from 1790 , when it was \(3.93\) million, until today, what would the population of the United States have been in 2000 ? For comparison, according to census data, the population of the United States in 2000 was \(281,421,906\). The population of the world was just over 6 billion people.

National health care spending: The following table shows national health care costs, measured in billions of dollars. $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Date } & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline \begin{array}{l} \text { Costs } \\ \text { in billions } \end{array} & 27.6 & 75.1 & 254.9 & 717.3 & 1358.5 \\ \hline \end{array} $$ a. Plot the data. Does it appear that the data on health care spending can be appropriately modeled by an exponential function? b. Find an exponential function that approximates the data for health care costs. c. By what percent per year were national health care costs increasing during the period from 1960 through 2000?

Sales growth: The total sales \(S\), in thousands of dollars, of a small firm is growing exponentially with time \(t\) (measured in years since the start of 2008). Analysis of the sales growth has given the following linear model for the natural logarithm of sales: $$ \ln S=0.049 t+2.230 \text {. } $$ a. Find an exponential model for sales. b. By what percentage do sales grow each year? c. Calculate \(S(6)\) and explain in practical terms what your answer means. d. When would you expect sales to reach a level of 12 thousand dollars?
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